Problem: Simplify and expand the following expression: $ \dfrac{3}{z - 2}+ \dfrac{4}{4z - 40}+ \dfrac{4}{z^2 - 12z + 20} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{4}{4z - 40} = \dfrac{4}{4(z - 10)}$ We can factor the quadratic in the third term: $ \dfrac{4}{z^2 - 12z + 20} = \dfrac{4}{(z - 2)(z - 10)}$ Now we have: $ \dfrac{3}{z - 2}+ \dfrac{4}{4(z - 10)}+ \dfrac{4}{(z - 2)(z - 10)} $ The least common multiple of the denominators is: $ (z - 2)(z - 10)$ In order to get the first term over $(z - 2)(z - 10)$ , multiply by $\dfrac{4(z - 10)}{4(z - 10)}$ $ \dfrac{3}{z - 2} \times \dfrac{4(z - 10)}{4(z - 10)} = \dfrac{12(z - 10)}{(z - 2)(z - 10)} $ In order to get the second term over $(z - 2)(z - 10)$ , multiply by $\dfrac{z - 2}{z - 2}$ $ \dfrac{4}{4(z - 10)} \times \dfrac{z - 2}{z - 2} = \dfrac{4(z - 2)}{(z - 2)(z - 10)} $ In order to get the third term over $(z - 2)(z - 10)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{4}{(z - 2)(z - 10)} \times \dfrac{4}{4} = \dfrac{16}{(z - 2)(z - 10)} $ Now we have: $ \dfrac{12(z - 10)}{(z - 2)(z - 10)} + \dfrac{4(z - 2)}{(z - 2)(z - 10)} + \dfrac{16}{(z - 2)(z - 10)} $ $ = \dfrac{ 12(z - 10) + 4(z - 2) + 16} {(z - 2)(z - 10)} $ Expand: $ = \dfrac{12z - 120 + 4z - 8 + 16}{4z^2 - 48z + 80} $ $ = \dfrac{16z - 112}{4z^2 - 48z + 80}$ Simplify: $ = \dfrac{4z - 28}{z^2 - 12z + 20}$